When \(\Sigma =\mathbf {I}_p\) , by Proposition REF , we know that \(\theta ^*(\delta )=(1-\kappa (\delta ))\theta _0\)  for some function \(\kappa \)  that only depends on \(\delta \) . In addition, based on equation (A.5) in Lemma A.6 of [1]}, we have \(\Vert \widehat{\Sigma }_n-\mathbf {I}_p\Vert =o(1)\)  and \(tr(\widehat{\Sigma }^{-1})=\Theta (p)\)  with probability tending to 1. Thus for \(\alpha =o(1)\) ,
\(\mathbb {E}_{\theta _0|\mathbf {X},\mathbf {y},\Sigma =\mathbf {I}_p}\Vert \theta ^*(\delta )-\mathbb {E}_{\theta _0|\mathbf {X},\mathbf {y},\Sigma =\mathbf {I}_p}[\theta ^*(\delta )]\Vert ^2&=& (1-\kappa (\delta ))^2 \mathbb {E}_{\theta _0|\mathbf {X},\mathbf {y},\Sigma =\mathbf {I}_p}\Vert \theta _0-\widehat{\theta }_{n,\alpha }\Vert ^2\\&=&(1-\kappa (\delta ))^2\frac{\sigma ^2}{n} tr\left((\widehat{\Sigma }_n+\alpha \mathbf {I}_p)^{-1} \right)\\&=&(1-\kappa (\delta ))^2\frac{\sigma ^2}{n} tr\left(\widehat{\Sigma }_n^{-1}-\alpha (\widehat{\Sigma }_n+\alpha \mathbf {I}_p)^{-1}\widehat{\Sigma }_{n}^{-1} \right)\\&=&\left(1+ O\left(\frac{1}{ \lambda _{\max }(\hat{\Sigma }_n)/\alpha +1}\right)\right)(1-\kappa (\delta ))^2\frac{\sigma ^2}{n} tr\left(\widehat{\Sigma }_n^{-1}\right)\\&=&(1+o(1))(1-\kappa (\delta ))^2\frac{\sigma ^2}{n} tr\left(\widehat{\Sigma }_n^{-1}\right).\) 
